De Morgans

7/31/2019 De Morgans

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Propositional logicThe propositional logic representlogic through proposition and

logical connectives.

We may define proposition as an

elementary atomic sentence that

may take either true value orfalse value but may not take

other value .


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Consider the following examples :

It is raining [it is a proposition as it

may either be true orfalse .

Australia won the ICCworld cup 2007.

It is a also proposition asit is true.

India is a continent It is a proposition as it isfalse

What did you eat It is not proposition as it

does not result in true orfalse.

How are you Not a proposition for the

similar reason as above


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A proposition is an

elementary atomic

sentence that may either

be true or false but may

not take other value .


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Proposition are also

called sentence orstatement after this

introduction , let us talkabout terms and symbol

used in propositionallogic .


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A symbol proposition is one thatdoes not contain any other

proposition of part . We will usethe lower case letter p, q, r., as

symbols for simple statements or

proposition .


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A compound proposition is one

with two or more simpleproposition as parts or what

We will call components . Acomponent of compound is any

whole proposition that is a parts

of larger proposition ; componentsmay themselves be compounds.


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For example , following arecompound proposition :

it is raining and wind is blowing .Take it or leave it .

If you work hard then you will

rewarded.


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An operators ( or connective ) joins simpleproposition into compounds, and joins

Compound into larger compounds. We will usethe symbols, +, ., and todesignate the sentential connectives . They arecalled sentential connectives because they join

sentences ( or what we are calling statement orpropositions ) . The symbol , ~ , is the onlyoperator that is not a connective ; it affectssingle statement only , and does not joinstatements into compounds.


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Different types of

connectives (oroperators ) used in

propositional logic are

given below :


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Represent by symbols + or v.

Disjunction means one of the

two arguments is true or bothe.g., p + q (or p v q) means p OR

q. It`s meaning is either p istrue , or q is true , or both


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Represent by symbols . or &

or . conjunction means both

arguments are true e.g., p . q(or p & q) means p and q. Its

meaning is both p and q aretrue .


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Represented x by symbols or

or

Implication means if one argument istrue then other argument is true e.g. , p

q(or p q or p q) means if p

then q . Its meaning is if p is true , then

q is true .


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Represented by symbols or .

Equivalence (or bi - conditional ) means

either both argument true or both false ,e.g., p q (or p q ) means if and

only if p is true then q is true . Its

meaning is p q are either both true or

both false .


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Not a connective actually , just anoperator . Represented by - or` or

~ (bar).

It is an operator that effects asingle statement only and does not

join two or more statement e.g., ~p (or p` or p ) means NOT P . Its

meaning is P Is False .


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A truth table is a

complete list ofpossible truth

values of a

proposition


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The not operators

works on singleproposition , thus it is

also called unaryconnective sometimes .


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If p denotes proposition , then

its negation will be denoted by~ p or q or p . If p is 0 (false),then ~ p is 1 (true ) and if p is 1

(true) then ~ p is 0 (false) . Thetruth table for this operation is

shown as follows:


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P P

0 1

1 0


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Also note that

NOT (NOT p) result into p itself

i.e.,

p = p

or (p) = por ~ (~p) = p.


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The OR connective workswith more then one

proposition . The compound p+ q has two (2) components

proposition (p and q), each ofwhich can be true or false .


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So there are (2* 2) possible

combination . The disjunctionof p with q (denoted as p + q

or p v q ) will be truewhenever p is true or q is

true or both are true .Consider the truth table given

below :


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p q p + q

0 0 0

0 1 1

1 0 1

1 1 1

TRUTH TABLE FOR DISJUNCTION (OR)


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The AND connective also

works with more than one

proposition compound The p q (or p & q) will be true

whenever both p and q aretrue


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p q p - q

0 0 0

0 1 0

1 0 0

1 1 1


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In the conditional p qthe first proposition (the if -

clause ) p here , is called the

antecedent and the second

proposition (then clause ) q

here , is called consequent.


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In the more complex conditionals

, the antecedent and theconsequent could themselves be

compound proposition . Theconditional p q will be false

when p is true and q is false . For

all other input combinations, itwill be true


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p q p q

0 0 1

0 1 1

1 0 0

1 1 1


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The conditional p q

may be expressed as follow:P q = p` + q


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A bi conditional results into false

when one of its component

proposition is true and other is false.


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That is, p q will

be 0 (false ) when p is 0and q is 1 or p is 1 and q

is 0 . For all other inputs ,p q is 1.


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p q p q

0 0 1

0 1 0

1 0 0

1 1 1


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The conditional p q may be

expressed as :p q = pq + p` . q`


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The proposition that

have some combinationof 1`s and 0`s in their

truth table columns,are called cotingencies


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The propositionshaving nothing but

1`s in their truth

table column, are

called tautologies.


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The propositions having

nothing but 0 `s in arecalled their truth table

column, are calledcontradictions


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Two statements are

consistent if and onlyif their conjunction is

not a contradiction .


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The converse of a conditional

proposition is determined by

interchanging the antecedentand consequent of given

conditional ..


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It results into new

conditional . e.g .,

converse of p q

is q p


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That is, if

p : it is raining.q : sky is not clear

.then , p q = if it

is raining then sky is not

clear .


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it`s converse will be new

conditional as givenbelow :

q p = if sky is not

clear then it is raining .


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The inverse of a conditional proposition

is another conditional having negated

antecedent and consequent . That is ,the inverse of p q is p` q` .

E.g., if

p : it is raining.q : sky is not clear .


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then , p q = if it is

raining then sky is not clear

it`s inverse will be a new

conditional as given below :

p` q` = if itis not raining then sky is not

clear .


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The cotnrapositive of a conditional isformed by creating another conditionalthat takes its antecedent as negated

consequent of earlier conditional andconsequent as antecedent of earlierconditional . That is , contrapositive

p q is ~q ~p or q por q` p`.


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1. 0 + P= P

0 . P = 0

Properties of 0

2. 1 + p = 11 . p = p

Properties of 1

3 . p + pq = p

p + (p + q)= p

absorption law .

4. p = p involution

5. p + p =pp . p = p

Idempotence law

6. p + p = 1p . p = 0

Complementaritylaw

7. p + q = q + pp . q = q . p

commulative law


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8. (p +q ) + r = p + (q + q)

(p . q ) . r = p . (q. r)

Associative Law

9.p . (q + r ) = (p .q) + (p .r)

p + (q . r ) = (p +q ). (p + r)p + p q = p + q

Distributive Law

10 . p + q = p . qp . q = p + q

De Morgan`s Law

11. p q = p + q Conditional

Elimination

12. p q = (p q ) . (q p ) Bi - ConditionalElimination


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Construction a truth table for the expression(A . ( A + B) ). What single term

Is the expression equivalent to ?

A B A + B (A . (A + B))

0 1 0 0

0 1 1 0

1 0 1 1

1 1 1 1


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Using truth table, prove p qis equivalent to ~ p ~ q

p q ~q ~ p p q ~ p ~q

0 0 1 1 1 1

0 1 0 1 1 1

1 0 1 0 0 0

1 1 0 0 1 1


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p q p p q p + q

0 0 1 1 1

O 1 1 1 1

1 0 0 0 0

1 1 0 1 1


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Prove that p q = q p

p q p q q p

0 0 1 1

0 1 0 0

1 0 0 0

1 1 1 1


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Prove that p q = (q p) . ( q p ).

P q p q p q q p (p q) . (q p)

0 0 1 1 1 1

0 1 0 1 0 0

1 0 0 0 1 0

1 1 1 1 1 1


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Consider some simple proposition

A : it is raining.B : wind is blowing .

C : I am not driving.


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1. A V B : it is raining and wind isblowing

2. ~ B : wind is not blowing .3. ~ A . C : It is not raining and I

am not driving ,

4. A. ~ c : it is raining and I amnot driving

5. A + B . C : it is raining or wind

is blowing and I amnot driving .

Prove that x + 1 is a tautology


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Prove that x + 1 is a tautology.

x 1 x + 1

0 1 1

1 1 1


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Prove that x + x ` is a tautology andx . X ` is a contradiction

x x` x + x` x . X`

0 1 1 0

1 0 1 o


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Before we start discussion abo utlogical operators, let us first

understand what a truth table is.

For example, following logicalstatements can have only one of the

two values (TRUE (YES) orFALSE (NO))


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1.I WANT TO HAVE

TEA.2.TEA IS READLY

AVABILABLE.


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Truth table is a table which

represents all the possiblevalue of logical variables /

statements along with all thepossible results of the given

combinations of values


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X Y R

1 1 1

1 0 0

0 1 0

0 0 0


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1Represent true value and 0

represent false value .

2This is a truth table i,e.,table of truth values of truth

functions.


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If result of any logical

statement or expression is

always TRUE or I , it iscalled Tautology and if the

result is always FALSE or 0it is called

FALLACY


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Now let us proceed with our

discussion about logical operators I,e.,

1. NOT operator2. OR operator3. AND operator

O O O


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NOT OPERATORthis operator operates onsingle variable and

operation performed by notoperator is called

complementation and thesymbol we use for it (bar) .

Thus x means complement of x


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Thus x means complement of x

and yz means complement of y z

. As we know, the variables used inBoolean equations have unique

characteristics that they may

assume only one of two possiblevalues 0 and 1 , where 0 denotes

FALSE and 1 denotes TRUE value. Thus the complement operation

can be defined quit simply .

0 = 1


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0 = 1

1 = 0

Truth table for not operator

x x

0 11 0

S l h b l l


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Several other symbol e.g ~ , are also usefor complementation symbol . If ~ is used

then ~ x is read is negation of x and ifsymbol is used then x is read complementof x .

x

x x


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Not operation is singular

or unary operation as itoperates on single variable .

Venn diagram for x is givenabove where shaded area

depicts x


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A second important

operator in Boolean algebrais or operator which denotes

operation called logical

addition and the symbol weuse for it is + .


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0 + 0 = 0

0 + 1 = 11 + 0 = 1

1 + 1 = 1

T th t bl f OR t


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Truth table for OR operator

X Y X + y0 0 0

0 1 11 0 1

1 1 1


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AND operator perform another

important operation of Boolean

algebra called logical multiplicationand the symbol for AND operation

is (.) dot . Thus x . Y will be

read as x AND Y . The rules forand operation are :


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0. 0 = 0

0 . 1 = 01. 0 = 0

1 . 1 = 0

T th t bl f AND t


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Truth table for AND operator

X Y X . Y

0 0 00 1 0

1 0 01 1 1

V di f X Y i i b l


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Venn diagram for X . Y is given belowwhere shaded area depicts (X .Y)

YX

Shaded portionshows

X . Y


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Logical variables are

combined by mean of logical

operation (AND, OR, NOT )

to form Boolean expression .

For example , x + y . Z + zis a Boolean expression .


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X Y Z0 0 0

0 0 1

0 1 0

0 1 1

1 0 0

1 0 1

1 1 0

1 1 1


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X Y Z Y . Z

0 0 0 0

0 0 1 0

0 1 0 0

0 1 1 1

1 0 0 0

1 0 1 0

1 1 0 0

1 1 1 1


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X Y Z Y . Z Y Z0 0 0 0 1

0 0 1 0 1

0 1 0 0 10 1 1 1 0

1 0 0 0 1

1 0 1 0 11 1 0 0 1

1 1 1 1 0


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X Y Z Y . Z YZ X + YZ

0 0 0 0 1 1

0 0 1 0 1 1

0 1 0 0 1 1

0 1 1 1 0 0

1 0 0 0 1 1

1 0 1 0 1 1

1 1 0 0 1 1

1 1 1 1 0 1


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X Y XY X + XY

0 0 0 0

0 1 0 0

1 0 0 11 1 1 1


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X Y X+Y (X+Y) X Y XY

0 0 0 1 1 1 10 1 1 0 1 0 0

1 0 1 0 0 1 01 1 1 0 0 0 0

PREPARE A TABLE OF COMBINATIONS FOR THE FOLLOWING BOOLEAN


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PREPARE A TABLE OF COMBINATIONS FOR THE FOLLOWING BOOLEANALGEBRA EXPRESSIONS.

1.XY+XY

2.XYZ+XYZ

3.XYZ+XY


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1. x y + x y is a 2 variable expression , its truth table asfollows :

X Y X Y X Y X Y X Y+ XY

0 0 1 1 1 0 1

0 1 1 0 0 1 1

1 0 0 1 0 0 0

1 1 0 0 0 0 0

b XYZ + X Y Z


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b. XYZ + X Y Z

Truth Table

X Y Z X Y Z XYZ X Y Z XYZ + X Y Z

0 0 0 1 1 1 0 0 0

0 0 1 1 1 0 0 1 1

0 1 0 1 0 1 0 0 0

0 1 1 1 0 0 0 0 0

1 0 0 0 1 1 0 0 0

1 0 1 0 1 0 0 0 0

1 1 0 0 0 1 1 0 1

1 1 1 0 0 0 0 0 0

c. X Y Z + X Y


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Truth TableX Y Z X Y Z X Y Z XY X Y Z + XY

0 0 0 1 1 1 0 0 0

0 0 1 1 1 0 0 0 0

0 1 0 1 0 1 1 0 1

0 1 1 1 0 0 0 0 0

1 0 0 0 1 1 0 1 11 0 1 0 1 0 0 1 1

1 1 0 0 0 1 0 0 0

1 1 1 0 0 0 0 0 0

A. X (Y + Z ) + X Y


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Truth Table

X Y Z Y Z ( Y +Z) X (Y + Z) XY X (Y + Z )+ XY0 0 0 1 1 1 0 0 0

0 0 1 1 0 1 0 0 0

0 1 0 0 1 1 0 0 0

0 1 1 0 0 0 0 0 0

1 0 0 1 1 1 1 1 1

1 0 1 1 0 1 1 1 1

1 1 0 0 1 1 1 0 1

1 1 1 0 0 0 0 0 0

B. XY (Z + YZ ) + Z Truth Table


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B. XY (Z YZ ) Z Truth Table

X Y Z Y Z YZ Z + YZ XY XY(Z +YZ) XY (Z + YZ )+ Z

0 0 0 1 1 0 0 0 0 1

0 0 1 1 0 0 1 0 0 0

0 1 0 0 1 1 1 0 0 1

0 1 1 0 0 0 1 0 0 0

1 0 0 1 1 0 0 1 0 1

1 0 1 1 0 0 1 1 1 1

1 1 0 0 1 1 1 0 0 1

1 1 1 0 0 0 1 0 0 0

C . A [ (B + C ) + C ] Truth Table


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C . A [ (B + C ) + C ] Truth Table

A B C B C ( B + C ) ( B + C) + C A [ (B + C) + C ]

0 0 0 1 1 1 1 0

0 0 1 1 0 1 1 0

0 1 0 0 1 0 1 0

0 1 1 0 0 1 1 0

1 0 0 1 1 1 1 1

1 0 1 1 0 1 1 1

1 1 0 0 1 0 1 1

1 1 1 0 0 1 1 1


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Gates are digital (two - state )circuits because the input and

output signals are either low

voltage ( denotes 0 ) or highvoltage (denotes 1 ). Gates are

often called logic circuits becausethey can be analyzed with Boolean

algebra


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A Gate is simply an

electronic circuit whichoperates on one or

more signals t o producean output signal .


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Inverter (NOT gate )

OR gateAND gate


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An inverter (NOT Gate ) is

a gate with only one inputsignal and one output

signal ; the output state isalways the opposite of the

input state


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An inverter is also called a NOT

gate because the output is not

the same as the input . The

output is sometimes called thecomplement (opposite) of the

input . Following tablessummaries the operation:


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X YLow HighHigh Low


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X Y

0 1

1 0

A low input I e 0 produce high output


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A low input I,e., 0 produce high output

I,e. 1, and vice versa. The symbol for

inverter is given in:

x x

NOT GATE SYMBOL


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The OR Gate has two or moreinput signals but only one

output signal . If any of theInput signals is 1 (high ). The

output signal is 1 (high)


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If all inputs are 0

then output is also 0. If one or more

inputs are 1, the

output is 1 TWO INPUT OR GATE


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X Y F

0 0 00 1 1

1 0 11 1 1

F = X + Y


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X Y Z F

0 0 0 0

0 0 1 10 1 0 1

0 1 1 1

1 0 0 1

1 0 1 1

1 1 0 1

1 1 1 1

F = X + Y + Z


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A

B

F

Two input OR gate


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AB FC

Three input OR gate


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A

B FC

Four input OR gate

D


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The AND Gate can have two ormore then two input signals

and produce an output signal .When all the input are 1 i.e,

that is high then output is 1

otherwise output is 0 only .


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X Y F0 0 0

0 1 01 0 0

1 1 1

F X Y


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X Y Z F

0 0 0 0

0 0 1 0

0 1 0 0

0 1 1 0

1 0 0 0

1 0 1 01 1 0 0

1 1 1 1

F = X Y Z


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2 Input AND gate

A

BF


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3 Input AND gate

A

B F

C


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4 Input AND gate

A

B FCD


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Boolean algebra being a system of

mathematics consists of fundamentallaws that are used to build a

workable cohesive framework uponwhich are based the theorems of

Boolean algebra . These fundamental

laws are known as Basic postulatesof Boolean


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1. if x = 0 then x = 1; and if x = 1 then x = 0

2. OR Relation s (Logical Addition)

0 + 0 = 0 00

0 + 1 = 1

1 + 0 = 1

1 + 1 = 1

OR

OR

OR

OR

0

1

1

1

01

1

0

11


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0 . 0 = 0

0 . 1 = 1

1 . 0 = 1

1 . 1 = 1

AND

AND

AND

AND

0

0

0

1

0

1

11

0

0

1

0


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0 = 1

1 = 0

0

01

1

NOT

NOT


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This is a very important principle used inBoolean algebra . This state s that startingwith Boolean relation can be derived by .

1. changing each OR sign ( + ) to an AND sign(.)

2. changing each AND sign ( . ) to an OR sign

(+)in3. Replacing each 0 by 1 and each 1 by 0 .

The derived relation


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using dualityprincipal is called

dual Of originalexpression.

For instance , we take postulate II related


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to logical addition , which states

A. 0 + 0 = 0B. 0 + 1 = 1

C. 1 + 0 = 1

D. 1 + 1 = 1

Now working according to above

d l h d d


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guidelines , + is changed to. And 0

s are replaced by 1s , these becomeA. 1 . 1 = 1

B. 1 . 0 = 0C. 0 . 1 = 0

D. 0 . 0 = 0


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1. 0 + x = x

2. 1 + x = 1

3. 0 . x = 0

1 x x

OR 00

0

OR 11

X

AND 1

0

X

AND X1

X


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0 X R

0 0 0

0 1 1


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1 X R

1 0 1

1 1 1


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0 X R

0 0 0

0 1 0


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1 X R

1 0 0

1 1 1


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OR

X

X

X

A. X + X = X i,e;

(gate representation for (a))

B. X . X = X i,e;


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,e;

AND XX

X

(gate representation for (b))


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X X R

0 0 01 1 1

0 + 0 = 01 + 1 = 1


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X X R

0 0 01 1 1

0 . 0 = 01 . 1 = 1


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(X) = X

X

X X = X


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X X R

0 1 0

1 0 1


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A. X + X = 1

OR

XX X + X = 1

( gate representation of (a)


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B. X . X = 0

ANDX

X X . X = 0

( gate representation of (b)


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X X X + X

0 1 1

1 0 1


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X X X . X

0 1 0

1 0 0


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A. X + Y = Y + X

ORX

Y

R = ORYXR

B. X . Y = Y . X


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ANDX

Y

R ANDYX

R

=

Truth Table X + Y = Y + X


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X Y X + Y Y + X

0 0 0 0

0 1 1 11 0 1 1

1 1 1 1

Truth Table for X . Y = Y . X


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X Y X . Y Y . X

0 0 0 0

0 1 0 01 0 0 0

1 1 0 1


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A. X + (Y + Z) = (X + Y ) + Z

ORYZ

X

OR =

R

Y + Z

ORXZ

Z

ORR

X + Y

B. X (YZ) = (X Y) Z


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( ) ( )

AND

Y

Z

X

AND =R

Y Z

ANDXY

Z

ANDR

X Y

Truth Table for X + (Y + Z) = (X + Y ) + Z


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X Y Z Y + Z X + Y X + (Y + Z ) (X + Y )+ Z

0 0 0 0 0 0 0

0 0 1 1 0 1 1

0 1 0 1 1 1 1

0 1 1 1 1 1 1

1 0 0 0 1 1 1

1 0 1 1 1 1 11 1 0 1 1 1 1

1 1 1 1 1 1 1


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A. X (Y+Z) = X Y+ XZ

OR

Y

Z

Y + Z

ANDX R

=

AND

AND

OR

X

Y

XY

XY

R

Z

B X +YZ = ( X +Y)( X+Z)


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B. X +YZ ( X +Y)( X+Z)

ANDY

Z

Y . Z

ORXR

=

OR

OR

AND

X

Y

X + Y

X+Y

R

Z

Truth Table for X + (Y + Z) = X Y + X Z

X Y Z Y Z X Y XZ X(Y Z ) XY XZ


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X Y Z Y + Z X Y XZ X(Y + Z ) XY + XZ

0 0 0 0 0 0 0 0

0 0 1 1 0 0 0 0

0 1 0 1 1 0 0 0

0 1 1 1 0 0 0 0

1 0 0 0 0 0 0 0

1 0 1 1 0 1 1 1

1 1 0 1 1 0 1 1

1 1 1 1 1 1 1 1


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A. X + XY = X

ANDY

Z

ORX Y

X

X

B X (X +Y) X


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B. X (X + Y) = X

ORY

ANDX+ Y

X

X

Truth Table for X + XY = X


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X Y XY X + XY

0 0 0 0

0 1 0 0

1 0 0 1

1 1 1 1

Column X and X + XY are identical . Henced Al i b d l b i ll


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proved . Also it can be proved algebraically

as

L. H.S = X + XY

= X (1 + Y)

Putting 1 + y = 1X . 1 = X = R.H.S

(b)Since rule (b) is dual of rule (a) it is also proved


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.

L.H.S. =X (X +Y) = X . X + XY

= X . X + XY

= X + XY= X ( 1 + Y )

= X . 1= X

= R. H.S

(X.X=X: Indempotence Law)

(using 1 + y =1 properties of 0 , 1)

(X .1 = X USING PROPERTIES OF0,1)

X XY X Y (Thi i h hi d di ib i l )


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X + XY = X + Y (This is the third distributive law )

AND

OR = OR

x

y

RRXY

y

X

NOT

X + XY = X + Y


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PROOF L.H.S = X + X Y= X . 1 + X Y= X (1 + Y ) + XY= X + XY + XY

= X + Y (X + X )= X + Y . 1= X + Y

R. H . S PROVED

(X. 1 = X PROPERTY OF 0 AND 1)

(1+Y= 1 PROPERTY OF 0 AND 1

(X +X =1 COMPLEMENTARITY LAW)

(Y .1 =Y property of 0 and 1)

1 0 + X = XPROPERTIES OF 0

2 0 . X = 0

3 1 + X = 1 PROPERTIES OF 1

4 1 X = X


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BOOLEAN ALGEBRA

RULES

4 1 . X = X

5 X + X = X INDEMPOTENCE LAW

6 X . X = X

7 X = X INVOLUTION

8 X + X = 1 COMPLEMENTARITY LAW

9 X . X = 0

10 X + Y = Y + X COMMULATIVE LAW

11 X . Y = Y.X

12 X + (Y + Z ) = (X + Y ) + Z ASSOCIATIVE LAW

13 X (YZ ) = (XY) Z

14 X (Y + Z ) = XY + XZ DISTRIBUTIVE LAW

15 X (Y + Z ) = (X + Y ) ( X + Z)

16 X + XY = X ABSORPTION LAW

17 X * ( X + Y ) = X

18 X + XY = X + Y

I h


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It states that x + y = x y

OR

ANDxy

R

NOT

X

Y

NOT

NOT

X

Y

R

PROOFTo prove this theorem , we need


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to recall complementarity laws, which

state thatX + X = 1 and X . X = 0

I,e, a logical variable / expression

when added with its complementproduces the output as 1 and when

multiplied with its complement

produces the output as 0 .

Now to prove DeMorgan s first theorem , we willuse complementarity laws .

let us assume that P = X + Y where , p , x , y


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are logical variables .

Then according to complementation law.P + P = 1 and P . P = 0

That means, if P , X , Y are boolean variablesthen this complementarity law must hold forvariables P . In other words , if p i.e X + Y = X Ythen

(x + y ) + x y must be equal to 1 .( x + y ) . X y must be equal to 0


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let us prove the first part , i, e.

(X + Y) + ( X Y ) = 1

( X + Y ) + X Y= ((X +Y ) + X) . ((X + Y ) + Y)= (X + X ) + Y) . (X + Y + Y)

= (1 + Y ) .( X + 1 )= 1 . 1= 1

(Ref. X + YZ = (X +Y ) (X + Z ))

(Ref . X + X = 1)(ref . 1 + x = 1)

So part is proved .Now let us proved second part I,e.


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( X + Y ) . X Y = 0( X + Y ) . X Y = X Y . ( X + Y)

= X Y X + X Y Y= X X Y + X Y Y= 0 . Y + X . 0= 0 + 0 = 0

So, second part is also proved , thus : X + Y = X Y

(ref X (YZ ) Z= (XY) Z)

(Ref . X (Y + Z )=XY + XZ )

(Ref . X .X = 0)

It t t th t


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It states that x . y = x + y

AND

ANDxy

R

NOT

XNOT

X

Y

RX . Y

NOT

Proof

If XY s complemrnt is X + Y then it must be true that

1. XY + ( X + Y ) = 1 AND


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2. XY ( X + Y ) = 0

L.H.S = XY + ( X + Y )= (X + Y + X ) . (X + Y + Y)= ( X + X + Y ) . ( X + Y + Y )

= ( 1 + Y ) . (X + 1 )= 1 . 1= 1 = R.H.S

(ref X + Y = Y + X )Ref.(X + Y )( X + Z ) = X + YZ)

ref X + X = 1ref 1 + x = 1

Now the second part i ,e


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L.H.S = XY . ( X + Y )= X Y X . X Y Y= X X Y . X Y Y

= 0 . Y + X . 0= 0 + 0 = 0 = R.H.SXY . ( X + Y) = 0 and XY + (X + Y ) = 1

ref X (Y + Z ) = XY + XZ

ref ( X . X = 0 )

XY . ( x + y ) = 0

Although the identifies above represent DeMorgan s theoremthe transformations more easily performed by following these step:


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y p y g p

1. complement the entire function2. change all the ANDs (.) to ORs (+) and all

the ORs (+) to ANDs (.)

3. complements each of the individual variables .

Break the line , charge the sign to demorganize aBoolean expression


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AB + A + AB = A + B + A + AB

= (A + B ) + ( A + AB )= ( A + B ) + (A + AB)

= A . B . (A . AB)= A . B ( A . ( A + B ))= AB (AA + AB)

= AB (0 + AB )= AB . 0 + ABAB = 0 + AABB= 0 + A A . 0= 0 + 0 = 0


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FOR EXAMPLE,X + X . Y = X

Its dual will be X . (X + Y ) = X

(Remember change .to + and viceversa; complement 0 and 1.)Similarly (X + Y) + Z = X + ( Y +Z )

is dual of (X . Y ) .Z = X . (Y . Z)X + 0 = X is dual of X . 1 = X

All boolean algebra is predicated on this two for one basis


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All boolean algebra is predicated on this two for one basis

Example : give dual of the following result in boolleansalgebra :

X . X = 0 for each X

Solution : using duality principal , dual of X . X = 0 isX + X = 1 (By changing (.) to ( +) and viseversa and by replacing 1s by 0s and viceversa)

Example : give the dual of X + 0 = X for each X .Solution : using duality principal , dual of X + 0 = X is

X .1 = X


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Example : state the principal of dulity in boolean algebra and give that dual of

the boolean expression :

(X + Y ). (X + Z ) . (Y + Z )Solution : principal of duality states that from every boolean relation, anotherboolean relation can be derived by

1. changing each OR sign ( +) to an AND ( .) sign2. changing each AND ( .) sign to an OR ( + ) sign3. replacing each 1 by o and each 0 by 1

The new derived relation is known as the dual of the original relation.Dual of ( X + Y) . ( X + Y ) . (Y + Z ) will be

(X . Y ) + ( X . Z ) + ( Y . Z ) = XY + X Z + Y Z


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De Morgans

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